Solve (5x-1)² > 0: Finding Positive Values of a Quadratic Function

To solve this problem, we'll follow these steps:

• Step 1: Recognize that any square of a real number is greater than zero unless the number itself is zero.
• Step 2: Find when the expression $5x - 1$ equals zero, since the square is zero only at this point.
• Step 3: Exclude this value to determine when the quadratic is strictly greater than zero.

Let's work through each step:
Step 1: The expression given is $y = (5x - 1)^2$. We know that the square of any non-zero real number is positive.
Step 2: Set the inner expression to zero to find the critical point:
$5x - 1 = 0$
Solving for $x$, we add 1 to both sides:
$5x = 1$
Divide both sides by 5:
$x = \frac{1}{5}$
Step 3: Therefore, $(5x - 1)^2 > 0$ for all $x \neq \frac{1}{5}$.
This means that the quadratic expression is greater than zero for all real values of $x$ except $x = \frac{1}{5}$.

Thus, the solution to the problem is $x\ne\frac{1}{5}$.